If the length of a closed organ pipe is $1.5 \, m$ and the velocity of sound is $330 \, m/s$,then the frequency for the second note is ... $Hz$.

$A$ source of frequency $340 \,Hz$ is kept above a vertical cylindrical tube closed at the lower end. The length of the tube is $120 \,cm$. Water is slowly poured in just enough to produce resonance. Then, the minimum height (velocity of sound $= 340 \,m/s$) of the water level in the tube for that resonance is (in $\,m$)

$A$ string of length $3 \ m$ and linear mass density $0.0025 \ kg/m$ is fixed at both ends. One of its resonance frequencies is $252 \ Hz$. The next higher resonance frequency is $336 \ Hz$. Then the fundamental frequency will be ..... $Hz$.

$A$ pipe's lower end is immersed in water such that the length of the air column from the top open end has a length of $25 \, cm$. The speed of sound in air is $350 \, m/s$. The air column is found to resonate with a tuning fork of frequency $1750 \, Hz$. By what minimum distance should the pipe be raised in order to make the air column resonate again with the same tuning fork?

$A$ pipe open at both ends produces a note of frequency $f_1$. When the pipe is kept with $3/4$th of its length in water,it produces a note of frequency $f_2$. The ratio $f_1/f_2$ is

The frequency of a vibrating air column in a pipe,open at both ends,is $f_1$. When $\frac{3}{4}$ of its length is immersed in water,the frequency of the vibrating air column is $f_2$. The value of $\frac{f_1}{f_2}$ is